Question

A particle moves along the x-axis so that at time t its position is given by s(t) = (t + 3)(t −1)3, t > 0.

For what values of t is the velocity of the particle increasing?

Answer 1

the change in position function s(t) w.r.t. time is the Velocity function , V(t)

V(t) = s ' (t)

Answer 2

Huh?

Answer 3

have you done derivatives,

Answer 4

yep

Answer 5

take the derivative of the position function, that will be the velocity function... if you take a second derivative, that is then the acceleration function

Answer 6

Velocity will be increasing when you have a net acceleration in the same direction you moving

Answer 7

So when t>1?

Answer 8

it is like the increasing /decreasing first derivative test from calc 1,

Answer 9

is that a cube on the (t - 1)^3 ?

Answer 10

Yes.

Answer 11

s(t) = (t + 3)(t −1)^3

v(t) = 4*(t - 1)^2 * (t+2)

a(t) = 12*(t - 1)*(t + 1)

position, velocity, acceleration functions

Answer 12

They ask when v(t) is increasing in value.


you can look at the acceleration function a(t) to decide this, since accel is the rate of change of velocity.

Answer 13

So the answer isn't t>1?

Answer 14

My choices are
0 < t < 1
t > 1
t > 0
The velocity is never increasing

Answer 15

oh i dont know yet

Answer 16

yeah looks like + acceleration and increased velocity when t is larger than 1

Answer 17

0 to 1, is negative values

Answer 18

Thank you. Can you also help me with another question?

Answer 19

SURE BRB

Answer 20

Okie. A balloon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the balloon is 9 feet above the ground. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35). Whenever you are ready. cx

Answer 21

What"s your attempt?

Answer 22

I have non. In all honesty, I'm just trying to finish a pretest as fast as I can with a not so shabby grade, so I can get back to my ACT prep.

Answer 23

none

Answer 24

yeah this related rate prob..

Answer 25

so youre just getting dan to answer your question for you? smh

Answer 26

It just looks pretty disrespectful to string him along like this, pretending to be following along while he gives you the answer when he's doing this for free out of the goodness of his own heart and all.

Answer 27

I'm also listening and taking into account what he has to say. You make it sound like I;m just about the answer, I do want to see how it is done because I need it, but at the moment I prefer a fast explained and concise answer for future reference. I apologize that my ACT is a bit more important to me. @Empty

Answer 28

Okay, why is 8 dh/dt?

Answer 29

rises at 8ft per second, that is the rate of change of height h, dh/dt

Answer 30

Ohhh, so what would be the next step? The picture helps a lot by the way, cx

Answer 31

you need something to relate the angle and the height, it is a right triangle

Answer 32

Do I just solve or a when h is 9?

Answer 33

THis is dynamic, the height is changing and the angle is changing at the same time always, they ask for how fast the angle is changing at that instant in time, when h=9...

Answer 34

you need some relationship between the angle and the height
then take the derivative with respect to time of everything, so you then have a relationship between the rates of change of height and angle...

Answer 35

I'm confused.

Answer 36

maybe the tangent will work...

tan(a) = h / 12

Answer 37

relationship between the angle and the height

Answer 38

DO I use 9 for h?

Answer 39

, that will tell you the angle 'a' at that instant... we need to get the rate the angle is changing

Answer 40

take the derivative of both sides of the function with respect to time

Answer 41

Of the tangent equation?

Answer 42

yes, need to use the chain rule or implicit differentiating

derivative of tan(a) will be first derivative with respect to angle a, then da/dt to give you d/dt of the function

Answer 43

tan(a) = h/12

\[\sec^2(a) * \frac{ da }{ dt } = \frac{ 1 }{ 12 }*\frac{ dh }{ dt }\]

Answer 44

I did it through and got 0.425. Does that sound correct>

Answer 45

i dont know

Answer 46

\[\frac{ da }{ dt }=\frac{ 1 }{ 12 }*\cos^2(a)*\frac{ dh }{ dt }\]


so tha tis the rate the angle a changes in terms of the rate h is changing

Answer 47

dh/dt is given at 8 ft/s

the angle 'a' needed is found from the instant when h=9 from the tangent
tan(a)=9/12
a = 0.6435

that is it, calculate the da/dt

Answer 48

yeah
da/dt = 0.43 rad / s
when the baloon is at h=9 ft

Answer 49

unless i f'ed up

Answer 50

YAY!!!

Answer 51

Do you know anything about Rolles Theorem?

Answer 52

all those probs are similar,

need a relationship between variables, and the problem usually has it for you...

derivative of both sides, and maybe some substitutions

Answer 53

maybe, is that one of those existence theorems... like intermediate value, mean value

Answer 54

Yep.

Answer 55

i remember the name, i forget what it says

Answer 56

Thats okay. Thank you for all the help. Have a great day and life. cx

Answer 57

k goodluck